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Pick an integer $n\ge 5$ and let $f\in C_{C}^{\infty}(\mathbb{R}^{N})$.

We want to use the Fourier transform to formally construct a function $u\in L^{\infty}(R^{n})$ that solves $\Delta^{2}u(x)=f(x)$.

For $u\in L^{1}(\mathbb{R}^{n})$, define

$\displaystyle\hat{u}=\mathcal{F}^{+}(u)(\omega)=\frac{1}{(2\pi)^{\frac{n}{2}}}\int_{\mathbb{R^{n}}}e^{-i\omega x}u(x)dx$ for $\omega\in\mathbb{R}^{n}$.

$\mathcal{F}^{+}:L^{1}(\mathbb{R}^{n})\to L^{\infty}(\mathbb{R}^{n})\implies\hat{u}\in L^{\infty}(\mathbb{R}^{n})$.

Taking the Fourier transform, we get that $16\pi^{2}|\omega|^{4}\hat{u}(\omega)=\hat{f}(\omega)$

Then $\displaystyle\hat{u}(\omega)=\frac{\hat{f}(\omega)}{16\pi^{2}|\omega|^{4}}$

So in principle,

$\displaystyle u(x)=\frac{1}{(2\pi)^{\frac{n}{2}}}\int_{\mathbb{R}^{n}}e^{i\omega x}\frac{\hat{f}(\omega)}{16\pi^{2}|\omega|^{4}}d\omega=\frac{1}{16\pi^{2}\cdot (2\pi)^{\frac{n}{2}}}\int_{\mathbb{R}^{n}}e^{i\omega x}\frac{\hat{f}(\omega)}{|\omega|^{4}}d\omega$

$\displaystyle =\frac{1}{16\pi^{2}\cdot (2\pi)^{\frac{n}{2}}}\int_{\mathbb{R}^{n}}e^{i\omega x}\hat{f}(\omega)(\int_{0}^{\infty}e^{-|\omega|^{4}\lambda}d\lambda)d\omega$

$\displaystyle =\frac{1}{16\pi^{2}\cdot (2\pi)^{\frac{n}{2}}}\int_{0}^{\infty}\int_{\mathbb{R}^{n}}e^{i\omega x}e^{-|\omega|^{4}\lambda}\hat{f}(\omega)d\omega d\lambda$

And I don't know where to go from here, or if I'm even going down the right lines.

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